An insurance agent randomly selected 10 of his clients and checked online price

Question

An insurance agent randomly selected 10 of his clients and checked online price quotes for their policies His summaries of the data are shown, where Diff is Local - Online. Test an appropriate hypothesis to see if there is evidence that drivers might save money by switching to an online agent. Use a = 0.05. Click here to view the insurance quote data. Identify the null and alternative hypotheses Choose Insurance Quote Data Are the conditions for matched pairs inference No. It cannot be determined if the Independent Yes, all three conditions for matched pairs in No, the Nearly Normal Condition is not sates No, the Independence Assumption is not sat No, the Paired Data Condition is not satisfied compute the test statistic. The test statistic is

Explanation / Answer

For null and alternative hypothesis,

Option B is correct because we need to test whether drivers can save money switching to online agent.

The Statcrunch output for this is:

Paired T hypothesis test:
D = 1 - 2 : Mean of the difference between Local and Online
H0 : D = 0
HA : D > 0
Hypothesis test results:

So,

Test statistic = 0.961

p - value = 0.1808

Fail to reject the null hypothesis. There is not enough evidence to conclude that drivers can save money switching to online agents.

Difference Mean Std. Err. DF T-Stat P-value Local - Online 48.6 50.573203 9 0.96098323 0.1808

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